Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given an interval $[a,b]$ that satisfies hypothesis of Rolle's theorem for function $$f(x) = x^4 + x^3 - x^2 + x -2$$

If $a = -2$, how do I find $b$ ?

This is what Ive done so far, Not sure if it is right...

Rolles theorem says if $f(a)=f(b)$ at some point in the interval $[a,b]$ the derivative of the function is zero.

so,$f(a)=32-8-8-4=12$

Lets find a $b$ where $f(b)=12$

so, $f(b)=b^4+b^3-b^2+b-2=12$

This is hard to do, but wolfram alpha says $b=1.77$

So somewhere between $-2$ and $1.77$ the derivative is zero.

share|improve this question
    
Yes. Not sure what your question is. You can also just double check your own work by taking the derivative $4x^3+3x^2-2x+1=0$, which again by wolfram alpha, we see the derivative is zero at $~-1.28$ which is between $-2$ and $1.77$ –  Daniel Montealegre Apr 2 '12 at 7:41
3  
I think you should check your calculations, $f(a) = f(-2) \neq 12$ ... –  in_wolframAlpha_we_trust Apr 2 '12 at 7:45
    
Once you have the correct $f(a)$, you might want to check $b=1$. –  Did Apr 2 '12 at 8:42

1 Answer 1

up vote 3 down vote accepted

Since $f$ can be rewritten as $f(x) = (x-1)(x+2)(x^2+1)$, you can see that $f(-2) = f(1)$. There is a local extremum somewhere in $(-2,1)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.